Abscissa of convergence for a very specific Dirichlet series / Euler product

Question

I am interested in the convergence of the following Euler product: $$ \prod_p \frac{1}{1-\chi(p)\cdot p^{-s}}. $$ The product is over all primes (in increasing order), with $\chi(p)=+1$ if $p \bmod 4 =3$ and $\chi(p)=-1$ if $p \bmod 4 =1$. Here $\chi(2)=\pm 1$, the sign does not matter. Also $s=\sigma+ it$ to use the standard notation.

Since the densities, for both set of primes, are identical, the signs will alternate nicely on average, and one would guess that the product converges not only for $\sigma>1$, but actually for $\sigma>0$. Is this the case, or is this still a conjecture?

Also, I'd love to see a plot of the orbit for a fixed value of $\sigma$, say $\sigma=0.75$. By orbit, I mean a scatterplot of the product in the complex plane (real vs. imaginary part), for a fixed value of $\sigma$, and $t$ varying between $0$ and (say) $1000$. See example here and here for a truncated (finite) product when $\chi$ is constant equal to $1$.The hole in the orbit shrinks to a single point (the origin) if you add all the primes [if RH if true; if not it shrinks to an empty set].

I'm working on a number theory tutorial, and I'd like to introduce the students to a situation where the orbit not only never crosses the origin (as for the standard Riemann Hypothesis if $0.5<\sigma<1$), but in addition, stays away and never gets too close to the origin. If possible, not a conjectured result, but an established one.

On a side not, I am interested to know what the abscissa of convergence is, for the associated Dirichlet series.

Update - Attempt with a different $\chi$

Let $p_k$ be the $k$-th prime ($p_1=2$ and so on) and $\chi(p_{2k})=-1$, $\chi(p_{2k+1})=+1$. Let $\chi$ be completely multiplicative. Could the abscissa of convergence be strictly less than $1$? If you look at the Euler product, say for $\sigma=0.75$, its convergence at $t=0$ is equivalent to the convergence (after taking the logarithm and standard asymptotic) of $$ \sum_{k=1}^\infty (-1)^{k+1}\frac{1}{p_k^\sigma}. $$ I believe that the Dirichlet test could positively answer the question. Actually, there was a question asked on MO eight years ago [see here], about a similar convergence problem (proved to converge) so similar that this one must also converge if $\sigma=0.75$, and indeed, as long as $\sigma>0$.

Assuming this is correct, and since convergence for a specific $\sigma_0=0.75$ at $t=0$ implies convergence for all $t$ for that particular $\sigma_0$, it implies that the abscissa of convergence is well below $1$. Does this also hold for the Euler product? If yes (and I assume that the answer is yes), then can the product still be equal to zero? The norm of the product can be very close to zero. My guess is that for the product to be zero if $\sigma<1$ (that is, to diverge) you would need $t=\infty$. I could be wrong.

Now defining the concept of hole. In dynamical systems, it's called a repulsion basin, usually not circular. Let $L(z,\chi)$ be the Dirichlet function represented by the product or its associated series. Here the hole is centered at the origin $z_0=(0,0)$ in the complex plane. It is the largest circle of radius $\rho$ such that $|L(z,\chi)-z_0|>\rho$ for all $z=\sigma+it$ in the complex plane, for a fixed value of $\sigma$, say $0.75$.

There is one thing where I am most certainly wrong: the hole is reduced to to a single point (the origin), contrarily to what I thought initially (a hole of radius $\rho>0$). But if you make a video of the orbit starting at $t=0$, no matter how fast your computer is, for a very, very long time the hole will be visible to the naked eye. It will shrink as $t$ increases, in the end to a single point (or worse, an empty set if there are some unexpected roots) but incredibly slowly. That's what it does for the standard RH case.

Final update: non-trivial examples with a hole

This does not contradict the universality property: this property is true if $\sigma<1$ and applies if $P$ is the set of all primes. Here $\sigma<1$ (typically but not necessarily) and $P$ can be a subset of primes, finite or infinite. A modified (generalized) universality property, in this context, would be: the orbit is dense between its internal boundary (the boundary of the hole), and its external boundary if bounded (e.g. if $\sigma>1$). Still have to prove it. Also, I assume that if there is a "thick" hole, there is only one.

I could not find "standard" examples (characters modulo $m$) with a hole if $P$ contains sufficiently many primes. I am pretty sure there are no such examples. So you need to look at cases where $P$ is infinite but sparse enough. It will work (still have to go through a round of double-checking to confirm this) if

$$\rho=\prod_{p\in P} \frac{1}{1+p^{-s}}>0$$

regardless of $\chi$. In other words, the above product needs to converge. Then the radius of the hole is $\geq \rho$. Using my generalized version of the universality property, I guess the radius is $\rho$ (exactly). This is compatible with RH ($\rho=0$) and GRH (in that case $P$ is too big anyway, and $\rho=0$).

Examples with a hole with $\rho>0$:

• If $P$ is the set of all primes that are the sum of two cubes.
• If $P$ is the set of twin primes ?? (this case is borderline)

Answer 1

What do you mean by the word "orbit"? Please define that term in the body of your question.

You ask where the product over $p$ converges. Although you did not specify it, I assume you mean product with terms in the order of increasing $p$. You need to specify the order of the terms if you're dealing with something not necessarily absolutely convergent.

Your $\chi$ is meant to be a totally multplicative function. For odd primes $p$, your $\chi(p)$ is $\lambda(p)\chi_4(p)$ where $\lambda$ is the Liouville function (the nonvanishing totally multiplicative cousin of the Moebius function that is $-1$ at all primes) and $\chi_4$ is the nontrivial Dirichlet character mod $4$, so a calculation with Euler products shows for ${\rm Re}(s) > 1$ that (as GH from MO points out) $$ \sum_{n \geq 1} \frac{\chi(n)}{n^s} = \frac{1}{1-\chi(2)/2^s}\frac{L(2s,\chi_4^2)}{L(s,\chi_4)} = \left(1 + \frac{\chi(2)}{2^s}\right)\frac{\zeta(2s)}{L(s,\chi_4)}. $$ The Euler product for $\zeta(2s)$ converges absolutely for ${\rm Re}(s) > 1/2$ and $1 + \chi(2)/2^s$ is nonvanishing for ${\rm Re}(s) > 0$, so for ${\rm Re}(s) > 0$ your Euler product will converge exactly where $\prod_p (1 - \chi_4(p)/p^s)$ converges.

The Generalized Riemann Hypothesis for $L(s,\chi_4)$ implies for ${\rm Re}(s) > 1/2$ that $L(s,\chi_4) \not= 0$ and that $L(s,\chi_4) = \prod_p 1/(1-\chi_4(p)/p^s)$. Conversely, the convergence and nonvanishing of $\prod_p 1/(1-\chi_4(p)/p^s)$ at a number $s_0$ with ${\rm Re}(s_0) > 1/2$ implies it equals $L(s_0,\chi_4)$ and that $L(s,\chi_4) \not= 0$ for ${\rm Re}(s) > {\rm Re}(s_0)$. So it is not reasonable to expect the product to converge for ${\rm Re}(s) < 1/2$ and it is reasonable to expect the product to converge for ${\rm Re}(s) \geq 1/2$ excluding at zeros of $L(s,\chi_4)$ on the critical line, but the only way you're going to have access to such results is through GRH.

It is utterly hopeless to expect a proof of convergence of either $\prod_p 1/(1-\chi_4(p)/p^{s_0})$ or $\sum \chi_4(p)/p^{s_0}$ at a single $s_0$ where ${\rm Re}(s_0) < 1$ unless you expect to be making dramatic progress towards GRH for $L(s,\chi_4)$, and the same is true with $\chi_4$ replaced by $\chi$. Concerning the Dirichlet series $\sum \chi(n)/n^s$, since $\chi(n) = \lambda(n)\chi_4(n)$ for odd $n$ it's reasonable to expect $\sum_{n \leq x} \chi(n) = O_\varepsilon(x^{1/2+ \varepsilon})$ if you want to use GRH for $L(s,\chi_4)$, so it's reasonable to expect that $\sum \chi(n)/n^s$ converges for ${\rm Re}(s) > 1/2$, but it's not reasonable to expect a proof of that without a major advance on GRH. For an analogous situation where a Dirichlet series was proved to converge on a half-plane bigger than its half-plane of absolute convergence by relying on a major result (modularity of elliptic curves over $\mathbf Q$), see my answer to an earlier MO question here.

Watch out: there is probably weird behavior for the Euler product on the critical line. We have $L(1/2,\chi_4) \not= 0$ (not weird) and if $\prod_p 1/(1-\chi_4(p)/\sqrt{p})$ has a nonzero value, then that value is $\sqrt{2}L(1/2,\chi_4)$ (weird!). See my paper "Partial Euler products on the critical line" or Kuo and Murty's paper "On a conjecture of Birch and Swinnerton-Dyer". Both appeared in Canadian J. Math. in 2005.

Since your purpose in working with the Euler product $\prod_p 1/(1-\chi(p)/p^s)$ is purely to have a concrete example for some students, I suggest you work with $\chi_4$, not $\chi$: the function $\chi_4$ is a more commonly encountered object (a Dirichlet character), $\chi_4$ doesn't have an artificial definition at $p = 2$, and the half-plane of convergence of the Dirichlet series $\sum \chi_4(n)/n^s$ is already known to be precisely ${\rm Re}(s) > 0$.

Answer 2

The OP's Euler product can be written as $$\bigl(1+\chi(2)2^{-s}\bigr)\frac{\zeta(2s)}{L(s,\chi_4)},$$ where $\chi_4$ is the nontrivial Dirichlet character mod $4$. As a result, for any $\sigma\in(1/2,1)$, the convergence of the OP's Euler product in the half-plane $\Re s>\sigma$ is equivalent to the non-vanishing of $L(s,\chi_4)$ in the same half-plane. In particular, the abscissa of convergence is less than $1$ if and only if a quasi Riemann Hypothesis holds for $L(s,\chi_4)$. The quasi Riemann Hypothesis is a major open problem for any Dirichlet $L$-function (or any automorphic $L$-function).